Yes, it's a matter of style, because you'd expect sizeof(char) to always be one.

On the other hand, it's very much an idiom to use sizeof(foo) when doing a malloc, and most importantly it makes the code self documenting.

Also better for maintenance, perhaps. If you were switching from char to wchar, you'd switch to

Copywchar *p = malloc( sizeof(wchar) * ( len + 1 ) );

without much thought. Whereas converting the statement char *p = malloc( len + 1 ); would require more thought. It's all about reducing mental overhead.

And as @Nyan suggests in a comment, you could also do

Copytype *p = malloc( sizeof(*p) * ( len + 1 ) );

for zero-terminated strings and

Copytype *p = malloc( sizeof(*p) * len ) );

for ordinary buffers.

Answer from brainjam on Stack Overflow
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Medium
csnotes.medium.com โ€บ malloc-in-c-for-int-and-char-c3677b857b65
Malloc in C, for int * and char * | by zihan | Medium
May 15, 2021 - (char *)malloc โ†’ I want to malloc to create an array of chars sizeof(char) โ†’ I want to create slots of chars, each slot must have the size of a char, i.e. 8 bits * (size + 1) โ†’ I want to create size slots of chars + 1 slot for the ...
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beginner - malloc and free of a char-array - Code Review Stack Exchange
I wrote a function to reverse a char-array (string). Since I'm beginner and didn't work with malloc and stuff before, maybe someone could take a look, if this is fine, what I'm doing here? char* More on codereview.stackexchange.com
๐ŸŒ codereview.stackexchange.com
February 9, 2017
Char, Malloc, and Struct Arrays - Stack Overflow
Im still learning C and I have a question regarding char arrays, malloc and structures. I have the following structure. Function prototype typedef struct example1{ char *name[20]; int ex_id; int More on stackoverflow.com
๐ŸŒ stackoverflow.com
c - malloc-ing a char array - Stack Overflow
For example; thisOne.name = X. The following; thisOne->name will work if thisOne is a pointer to an instance of arr in memory. Okay. Based on your implementation, it seems like the call; ... is allocating memory of 128 characters (128 * 8 bits) into memory. This is fine, if the structure is like so; ... However your structure is in fact declaring an array of 128 char pointers, which is conflicting with your malloc ... More on stackoverflow.com
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c - Memory allocation for char array - Stack Overflow
If you're stuck with the archaic C89/C90 standard, then you can only define variables at the start of a block, and arrays have sizes known at compile time, so you have to use dynamic memory allocation โ€” malloc(), free() etc. ... First declare a pointer to a "char". More on stackoverflow.com
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September 7, 2015
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Arduino Forum
forum.arduino.cc โ€บ projects โ€บ programming
Dynamically Allocate Memory For An Array Of Chars - Programming - Arduino Forum
December 29, 2022 - I'm trying to create dynamic array of strings, but obviously I'm missing something... This does not compile: char **AOC; AOC = malloc(20 * sizeof(char *)); AOC[0] = malloc(20 * sizeof(char)); with this error: AOC = maโ€ฆ
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Quora
quora.com โ€บ How-do-I-dynamically-allocate-a-char-array-using-a-malloc-function-in-C
How to dynamically allocate a char array using a malloc function in C - Quora
In order to allocate memory dynamically using C language the following code will be enough: char *s = (char *)malloc(20 * sizeof(char)); The above line allocates memory for storing 20 characters or in fact 19 ch...
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Wikibooks
en.wikibooks.org โ€บ wiki โ€บ A_Little_C_Primer โ€บ C_Dynamic_Memory_Allocation_&_Deallocation
A Little C Primer/C Dynamic Memory Allocation & Deallocation - Wikibooks, open books for an open world
September 10, 2008 - /*malloc.c */ #include <stdio.h> #include <stdlib.h> /*For "malloc", "exit" functions. */ int main() { char *p; /*Pointer to array. */ unsigned count; /*Size of array. */ puts( "Size of array?" ); scanf( "%d", &count ); /*Get size in bytes. */ p = malloc( (size_t)count ); /*Allocate array.
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Eskimo
eskimo.com โ€บ ~scs โ€บ cclass โ€บ notes โ€บ sx11a.html
11.1 Allocating Memory with malloc
The 100 boxes in the figure don't have a name next to them, because they're not storage for a variable we've declared.) As a second example, we might have occasion to allocate a piece of memory, and to copy a string into it with strcpy: char *p = malloc(15); /* incomplete -- malloc's return ...
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CopyProgramming
copyprogramming.com โ€บ howto โ€บ how-to-malloc-a-char-array-in-c
Char: Creating a char array using malloc in the C programming language
May 8, 2023 - Memory - malloc vs array in C, What was taught is that malloc (10*sizeof (char)) allocates enough bytes on the heap to store 10 characters and returns a pointer to the first byte which can be saved in another variable as follows char *x = malloc (10*sizeof (char)).
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The C Book
publications.gbdirect.co.uk โ€บ c_book โ€บ chapter5 โ€บ sizeof_and_malloc.html
The C Book โ€” Sizeof and storage allocation
As a first illustration of the use of malloc, here's a program which reads up to MAXSTRING strings from its input and sort them into alphabetical order using the library strcmp routine. The strings are terminated by a โ€˜\nโ€™ character. The sort is done by keeping an array of pointers to the strings and simply exchanging the pointers until the order is correct. This saves having to copy the strings themselves, which improves the efficency somewhat. The example is done first using fixed size arrays, then another version uses malloc and allocates space for the strings at run time.
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LabEx
labex.io โ€บ tutorials โ€บ c-how-to-manage-memory-for-char-types-in-c-510337
How to manage memory for char types in C | LabEx
char *dynamic_char = malloc(sizeof(char)); // Heap allocation *dynamic_char = 'B'; // Always free dynamically allocated memory free(dynamic_char); ... char string[10] = "LabEx"; // Static character array char *dynamic_string = malloc(10 * ...
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4

Here are some things that may help you improve your code.

Fix the bug

Once memory is freed, it should not be referenced again. Unfortunately, your code allocates memory and then frees it and then returns a pointer to the freed memory. That's a serious bug! To fix it, simply omit the free within the function and make sure the caller calls free instead. Alternatively, you could avoid all of that by reversing the passed string in place.

Use the required #includes

The code uses strlen which means that it should #include <string.h> and malloc and free which means that it should #include <stdlib.h>. It was not difficult to infer, but it helps reviewers if the code is complete.

Use const where practical

In your revere_string routine, the string passed into the function is not and should not be altered. You should indicate that fact by declaring it like this:

char* reverse_string(const char* string)

Check for NULL pointers

The code must avoid dereferencing a NULL pointer if the call to malloc fails. The only indication that it has failed is if malloc returns NULL; if it does, it would probably make most sense to immediately return that NULL pointer.

Learn to use pointers instead of indexing

Using pointers effectively is an important C programming skill. This code could be made much simpler by doing an in-place reversal of the passed string and by using pointers:

char* reverse_string(char* string) {
    if (string == NULL) 
        return string;
    char *fwd = string;
    char *rev = &string[strlen(string)-1];

    while (rev > fwd) {
        char tmp = *rev;
        *rev-- = *fwd;
        *fwd++ = tmp;
    }
    return string;
}
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CopyProgramming
copyprogramming.com โ€บ howto โ€บ malloc-and-free-of-a-char-array
Beginner: Char-array Memory Allocation and Deallocation Using Malloc and Free
May 30, 2023 - C - Malloc on char *arr[], If you need N strings, you need to allocate N pointers to char each able to contain MAX_SIZE characters. First you need to allocate the array: char ** buff = malloc (N*sizeof (char *)); Please note how the result of the malloc is assigned to buff and not to *buff ...
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Cprogramming
cboard.cprogramming.com โ€บ c-programming โ€บ 159741-setting-character-malloced-char-array.html
Setting character in malloc'ed char array
#include <stdio.h> #include <stdlib.h> int main (int argc, char const* argv[]) { char str_one[4092] = "This is string number one"; char * str_two = "This is string number two"; char * str_three; str_three = (char *) malloc(4092); str_three = "This is the third string"; str_one[5] = 'X'; printf("%s\n", str_one); str_two++; *str_two = 'X'; // segfaults here str_two--; printf("%s\n", str_two); str_three++; *str_three = 'X'; // segfaults here str_three--; printf("%s\n", str_three); return 0; }
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Stack Overflow
stackoverflow.com โ€บ questions โ€บ 22390713 โ€บ char-malloc-and-struct-arrays
Char, Malloc, and Struct Arrays - Stack Overflow
for (i=0;i<20;i++) { info->name[i]=(char *)malloc(sizeof(char)*20); } Also, you are treating your struct pointer info as an array:
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Diveintosystems
diveintosystems.org โ€บ book โ€บ C2-C_depth โ€บ strings.html
2.6.1. C's Support for Statically Allocated Strings (Arrays ...
When dynamically allocating space to store a string, itโ€™s important to remember to allocate space in the array for the terminating '\0' character at the end of the string. The following example program demonstrates static and dynamically allocated strings (note the value passed to malloc):
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48

Declaring Static Character Arrays (strings)

When you know (or have a reasonable idea how large your array needs to be, you can simply declare an array of sufficient size to handle your input (i.e. if you names are no longer than 25 characters, then you could safely declare name[26]. For strings, you always need at minimum the number of chars to store + 1 (for the null-terminating character).

If there may be a few characters more than 25, there's nothing wrong with declaring your array few bytes longer than needed to protect against accidental writing beyond the end of the array. Say name[32].

Let's declare an array of 5-characters below and look at how the information is stored in memory.

char name[5] = {0}; /* always initialize your arrays */

The above declaration creates an array of 5-contiguous bytes on the stack for your use. e.g., you can visualize the 5-bytes of memory initialized to zero as follows:

        +---+---+---+---+---+
 name   | 0 | 0 | 0 | 0 | 0 |
        +---+---+---+---+---+
          \
           'name' holds the starting address for
            the block of memory that is the array
            (i.e. the address of the first element
             like: 0x7fff050bf3d0)

Note: when using name to hold a 'character string', the actual string can be no longer than 4 chars because you must end a sting with a null-terminating character, which is the null-character '\0' (or simply numeric 0, both are equivalent)

To store information in name, you can either do it by assigning characters one-at-a-time:

name[0] = 'J';
name[1] = 'o';
name[2] = 'h';
name[3] = 'n';
name[4] =  0;   /* null-terminate. note: this was already done by initialization */

In memory you now have:

        +---+---+---+---+---+
 name   | J | o | h | n | 0 |    /* you actually have the ASCII value for */
        +---+---+---+---+---+    /* each letter stored in the elements    */

Of course, nobody assigns one character at a time in this manner. You options are many, using one of the functions provided by C, e.g. strcpy, strncpy, memcpy, or by reading information from a file stream, or file descriptor with fgets, getline, or by using a simple loop and index variable to do the assignment, or by using one of the string formatting functions, e.g. sprintf, etc... For example you can accomplish the same thing with any of the following:

/* direct copy */
strcpy (name, "John");
strncpy (name, "John", 5);
memcpy (name, "John", sizeof "John");  /* include copy of the terminating char */

/* read from stdin into name */
printf ("Please enter a name (4 char max): ");
scanf ("%[^\n]%*c", name);

Note: above with strncpy, if you had NOT initialized all element to 0 (the last of which will serve as your null-terminating character, and then used strncpy (name, "John", 4); you would need to manually terminate the string with name[4] = 0;, otherwise you would not have a valid string (you would have an unterminated array of chars which would lead to undefined behavior if you used name where a string was expected.)

If you do not explicitly understand this STOP, go read and understand what a null-terminated string is and how it differs from an array of characters. Seriously, stop now and go learn, it is that fundamental to C. (if it doesn't end with a null-terminating character - it isn't a c-string.

What if I don't know how many characters I need to store?

Dynamic Allocations of Character Strings

When you do not know how many characters you need to store (or generally how many of whatever data type), the normal approach is to declare a pointer to type, and then allocate a reasonably anticipated amount of memory (just based on your best understanding of what you are dealing with), and then reallocate to add additional memory as required. There is no magic to it, it is just a different way of telling the compiler how to manage the memory. Just remember, when you allocate the memory, you own it. You are responsible for (1) preserving a pointer to the beginning address of the memory block (so it can be freed later); and (2) freeing the memory when you are done with it.

A simple example will help. Most of the memory allocation/free functions are declared in stdlib.h.

char *name = NULL;  /* declare a pointer, and initialize to NULL */

name = malloc (5 * sizeof *name); /* allocate a 5-byte block of memory for name */

if (!name) {    /* validate memory was allocated -- every time */
    fputs ("error: name allocation failed, exiting.", stderr);
    exit (EXIT_FAILURE);
}

/* Now use name, just as you would the statically declared name above */
strncpy (name, "John", 5);

printf (" name contains: %s\n", name);

free (name);    /* free memory when no longer needed.
                   (if reusing name, set 'name = NULL;') 
                 */

Note: malloc does NOT initialize the contents of the memory it allocates. If you want to initialize your new block of memory with zero (as we did with the static array), then use calloc instead of malloc. You can also use malloc and then call memset as well.

What happens if I allocate memory, then need More?

As mentioned above discussing dynamic memory, the general scheme is to allocate a reasonable anticipated amount, then realloc as required. You use realloc to reallocate the original block of memory created by malloc. realloc essentially creates a new block of memory, copies the memory from your old block to the new, and then frees the old block of memory. Since the old block of memory is freed, you want to use a temporary pointer for reallocation. If reallocation fails, you still have your original block of memory available to you.

You are free to add as little or as much memory as you like at any call to realloc. The standard scheme usually seen is to start with some initial allocation, then reallocate twice that amount each time you run out. (the means you need to keep track of how much memory is currently allocated).

To sew this up, let's end with a simple example that simply reads a string of any length as the first argument to the program (use "quotes" if your string contains whitespace). It will then allocates space to hold the string, then reallocate to append more text to the end of the original string. Finally it will free all memory in use before exit:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main (int argc, char **argv) {

    if (argc < 2) { /* validate input */
        fprintf (stderr, "error: insufficient input.  usage: %s \"name\"\n",
                 argv[0]);
        return 1;
    }

    size_t len = strlen (argv[1]);  /* length of input  */
    size_t sz_mem = len + 1;        /* memory required  */

    char *name = malloc (sz_mem * sizeof *name);  /* allocate memory for name */

    if (!name) {    /* validate memory created successfully or throw error */
        fputs ("error: name allocation failed, exiting.", stderr);
        return 1;
    }

    printf ("\n allocated %zu bytes of memory for 'name'\n", sz_mem);
    memset (name, 0, sz_mem); /* initialize memory to zero (optional) */

    strncpy (name, argv[1], sz_mem);  /* copy the null-terminator as well */
    printf (" name: '%s' (begins at address: %p)\n", name, name);

    /* realloc - make name twice as big */
    void *tmp = realloc (name, 2 * sz_mem);  /* use a temporary pointer */
    if (!tmp) {                              /* check realloc succeeded */
        fprintf (stderr, "error: virtual memory exhausted, realloc 'name'\n");
        return 1;
    }
    memset (tmp + sz_mem, 0, sz_mem * sizeof *name); /* zero new memory */
    name = tmp;         /* assign new block to name       */
    sz_mem += sz_mem;   /* update current allocation size */

    printf (" reallocated 'name' to %zu bytes\n", sz_mem);
    strncat (name, " reallocated", sizeof " reallocated");

    printf ("\n final name : '%s'\n\n", name);

    free (name);

    return 0;
}

Use/Output

$ ./bin/arraybasics "John Q. Public"

 allocated 15 bytes of memory for 'name'
 name: 'John Q. Public' (begins at address: 0xf17010)
 reallocated 'name' to 30 bytes

 final name : 'John Q. Public reallocated'

Memory Check

When you dynamically allocate memory, it is up to you to validate you are using the memory correctly and that you track and free all the memory you allocate. Use a memory error checker like valgrind to veryify your memory use is correct. (there is no excuse not to, it is dead-bang-simple to do) Just type valgrind yourprogramexe

$ valgrind ./bin/arraybasics "John Q. Public"
==19613== Memcheck, a memory error detector
==19613== Copyright (C) 2002-2012, and GNU GPL'd, by Julian Seward et al.
==19613== Using Valgrind-3.8.1 and LibVEX; rerun with -h for copyright info
==19613== Command: ./bin/arraybasics John\ Q.\ Public
==19613==

 allocated 15 bytes of memory for 'name'
 name: 'John Q. Public' (begins at address: 0x51e0040)
 reallocated 'name' to 30 bytes

 final name : 'John Q. Public reallocated'

==19613==
==19613== HEAP SUMMARY:
==19613==     in use at exit: 0 bytes in 0 blocks
==19613==   total heap usage: 2 allocs, 2 frees, 45 bytes allocated
==19613==
==19613== All heap blocks were freed -- no leaks are possible
==19613==
==19613== For counts of detected and suppressed errors, rerun with: -v
==19613== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 2 from 2)

In the output, the following lines are of particular significance:

==19613== HEAP SUMMARY:
==19613==     in use at exit: 0 bytes in 0 blocks
==19613==   total heap usage: 2 allocs, 2 frees, 45 bytes allocated

This tells you that all memory allocated during your program has been properly freed. (make sure you close all open file streams, they are dynamically allocated as well).

Of equal importance is the ERROR SUMMARY:

==19613== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 2 from 2)

There are no errors in the memory use. If you attempt to read, write or free memory from a location outside your block, or from an unitialized location or that would leave other memory unreachable, that information will show as an error.

(the suppressed: 2 from 2 just relate to additional debug libraries not present on my system)

This ended up longer than intended, but if it helps, it was worth it. Good luck.

2 of 5
5

How can I announce the array and then define it's size?

Don't; 'announce' it when you know what size it needs to be. You can't use it before then anyway.

In C99 and later, you can define variables when needed โ€” anywhere in a statement block. You can also use VLAs (variable-length arrays) where the size is not known until runtime. Don't create enormous arrays as VLAs (e.g. 1 MiB or more โ€” but tune the limit to suit your machine and prejudices); use dynamic memory allocation after all.

If you're stuck with the archaic C89/C90 standard, then you can only define variables at the start of a block, and arrays have sizes known at compile time, so you have to use dynamic memory allocation โ€” malloc(), free() etc.

๐ŸŒ
George Washington University
www2.seas.gwu.edu โ€บ ~simhaweb โ€บ C โ€บ modules โ€บ module3 โ€บ module3.html
Module 3: Pointers, strings, arrays, malloc
Strings are enclosed in double-quotes. ... Declare an array with a static size. Declare an array variable without a size, and allocate memory dynamically. ... // Note the use of the "sizeof" keyword. A2 = (int*) malloc (sizeof(int) * 10); for (i=0; i < 10; i++) { A2[i] = i * 100;